Re: MOT supply

["Tesla list" <tesla-at-pupman-dot-com>]

Original poster: "Gregory Hunter by way of Terry Fritz <twftesla-at-qwest-dot-net>" <ghunter31014-at-yahoo-dot-com>

Hi Matt,

No need for confusion. The two MOTs put out about
4kvac. The doublers shift this up to 8kvac (or about
12kvdc). Imagine a 60Hz sine wave with a horizontal
line drawn right though the middle, dividing the wave
into positive and negative halves ("peaks" are
positive, "valleys" are negative). Now throw away
either all the "peaks" leaving only the "valleys", or
else throw away all the "valleys", leaving only the
"peaks". You end up with a train of pulses separated
by a blank space. This is what the unloaded output of
the voltage doubled MOT supply looks like. You get a
train of 12kvdc pulses at 60Hz, with a zero volts
blank space between each one. The pulses can be all
positive or all negative, depending on which way the
diodes are pointed. BTW, polarity doesn't make a bit
of difference as far as performance goes. If the two
MOTs are ballasted down to 1800VA, then the average
output of the supply will be 1800VA. However, during
the "missing" half-cycle the MOTs are actually
recharging the doubler caps, so when a pulse does come
out of the supply, it contains the stored energy from
the "missing" half cycle as well as the energy of the
present half cycle, so to the Tesla coil tank circuit
it momentarily "feels" like 3600VA.

Regards,

Greg

--- Tesla list <tesla-at-pupman-dot-com> wrote:
> Original poster: "by way of Terry Fritz
> <twftesla-at-qwest-dot-net>" <A123X-at-aol-dot-com>
> 
> 
> Well I'm still pretty confused as to what to do. I
> need to get a few things
> straight. If I use two MOT's in series and then the
> voltage doubler I get
> around 8Kv pulsed for half of every cycle. If 1800Va
> is put out of the two
> MOT's then will I be getting 3600watts each pulse or
> will it stay 1800watts for
> each pulse? 
> 
> Mark
> 
> 


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